Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10100;
const int MAX_BIT = 32;
int n;
int x[MAXN], y[MAXN];
vector<pair<int, int> > a, b;
void read() {
  int i;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%d", &x[i]);
  }
  for (i = 0; i < n; ++i) {
    scanf("%d", &y[i]);
  }
}
void gauss(int a[], vector<pair<int, int> > &v) {
  int k = 0;
  for (int i = MAX_BIT - 1; i >= 0; --i) {
    int idx = -1;
    for (int j = k; j < n; ++j) {
      if ((a[j] & (1 << i))) {
        idx = j;
        break;
      }
    }
    if (idx == -1) continue;
    if (idx != k) {
      swap(a[k], a[idx]);
      v.push_back(make_pair(k, idx));
      v.push_back(make_pair(idx, k));
      v.push_back(make_pair(k, idx));
    }
    for (int j = 0; j < n; ++j) {
      if (k != j && (a[j] & (1 << i))) {
        a[j] ^= a[k];
        v.push_back(make_pair(j, k));
      }
    }
    k++;
  }
}
void solve() {
  int i, j;
  gauss(x, a);
  gauss(y, b);
  for (i = 0; i < MAX_BIT; ++i) {
    int bit = -1;
    for (j = MAX_BIT - 1; j >= 0; --j) {
      if ((x[i] & (1 << j))) {
        bit = j;
        break;
      }
    }
    if (bit == -1) continue;
    for (j = 0; j < MAX_BIT; ++j) {
      if ((x[j] & (1 << bit)) ^ (y[j] & (1 << bit))) {
        x[j] ^= x[i];
        a.push_back(make_pair(j, i));
      }
    }
  }
  for (i = 0; i < n; ++i) {
    if (x[i] != y[i]) {
      printf("-1\n");
      return;
    }
  }
  printf("%d\n", (int)a.size() + (int)b.size());
  for (i = 0; i < (int)a.size(); ++i) {
    printf("%d %d\n", a[i].first + 1, a[i].second + 1);
  }
  for (i = (int)b.size() - 1; i >= 0; --i) {
    printf("%d %d\n", b[i].first + 1, b[i].second + 1);
  }
}
int main() {
  read();
  solve();
  return 0;
}